You have found the following ages (in years) of all 4 porcupines at your local zoo: $ 10,\enspace 1,\enspace 6,\enspace 10$ What is the average age of the porcupines at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 4 porcupines at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{10 + 1 + 6 + 10}{{4}} = {6.8\text{ years old}} $ Find the squared deviations from the mean for each porcupine. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $10$ years $3.2$ years $10.24$ years $^2$ $1$ year $-5.8$ years $33.64$ years $^2$ $6$ years $-0.8$ years $0.64$ years $^2$ $10$ years $3.2$ years $10.24$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{10.24} + {33.64} + {0.64} + {10.24}} {{4}} $ $ {\sigma^2} = \dfrac{{54.76}}{{4}} = {13.69\text{ years}^2} $ The average porcupine at the zoo is 6.8 years old. The population variance is 13.69 years $^2$.